If a man has cystic fibrosis and his wife is a carrier of the gene, what is the likehood of their offspring having the disease ? Use a punnet square to justify your answer

Use the Punnett square.

https://www.google.com/search?client=safari&rls=en&q=punnett+square&ie=UTF-8&oe=UTF-8&gws_rd=ssl

Remember that it is a recessive gene.

cross red bull with a roan cow

To determine the likelihood of their offspring having cystic fibrosis, we can use a Punnett square. First, let's define the genotypes of the parents:

The man with cystic fibrosis would have the genotype "ff" since cystic fibrosis is an autosomal recessive genetic disorder. The lowercase "f" represents the mutated gene.

The wife, as a carrier of the gene, would have the genotype "Ff" where the uppercase "F" represents the normal gene and the lowercase "f" represents the mutated gene.

Now, let's create a Punnett square to determine the possible genotypes and the likelihood of their offspring having cystic fibrosis:

```
| F f
______|________________
F | FF Ff
f | Ff ff
```

When we cross the genotypes "FF" and "Ff," none of their offspring will have cystic fibrosis since there are no lowercase "f" alleles present. If both parents were carriers (Ff x Ff), we would use the Punnett square to determine that 25% of their offspring will have cystic fibrosis (ff).

Therefore, if the man has cystic fibrosis (ff) and the wife is a carrier (Ff), there is a zero percent chance that their offspring will have the disease.